$area \: of \: \triangle = \frac{1}{2} \times base \times altitude \\ \\ 40 = \frac{1}{2} \times x \times (x + 2) \\ \\ 80 = {x }^{2} +2x \\ \\ {x}^{2} + 2x - 80 = 0 \\ {x}^{2} + 10x - 8x - 80 \\ x(x + 10) - 8(x + 10) = 0\\ (x + 10)(x - 8) = 0 \\ x + 10 = 0 \: \: or \: \: x - 8 = 0 \\ x = - 10 \: \: or \: \: x = 8 \\ base \: can \: not \: be \: - ve \\ x \neq - 10 \\ x = 8 \\ x + 2 = 8 + 2 = 10 \\$

Therefore

Altitude = 10 m

Base = 8 m

8 0

3 years ago

A rectangle is placed around a semi circle as shown below. The width of the rectangle is 7yd. find the area of the shaded region

Nady [450]

Answer:

21.07 yd^2

Step-by-step explanation:

The width of the rectangle is also the radius of the semicircle. The length of the rectangle is 2 radii, or 14 yd.

The area of the shaded region is the same as the area of the semicircle subtracted from the area of the rectangle.

area of shaded region = area of rectangle - area of semicircle

A = LW - (1/2)(pi)r^2

A = 14 yd * 7 yd - (1/2)(3.14)(7 yd)^2

A = 98 yd^2 - (1.57)(49 yd^2)

A = 98 yd^2 - 76.93 yd^2

A = 21.07 yd^2

4 0

2 years ago